Consider the initial value problemy= t2 + ey, y(0) = 0.

Chapter 8, Problem 2

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Consider the initial value problemy= t2 + ey, y(0) = 0. (i)Using the RungeKutta method with step size h, we obtain the results in Table 8.6.5.These results suggest that the solution has a vertical asymptote between t = 0.9 and t = 1.0(a) Let y = (t) be the solution of initial value problem (i). Further, let y = 1(t) be thesolution ofy= 1 + ey, y(0) = 0, (ii)and let y = 2(t) be the solution ofy= ey, y(0) = 0. (iii)Show that2(t) (t) 1(t) (iv)on some interval, contained in 0 t 1, where all three solutions exist.(b) Determine 1(t) and 2(t). Then show that (t) for some t betweent = ln 2 = 0.69315 and t = 1.(c) Solve the differential equations y= ey and y= 1 + ey, respectively, with the initialcondition y(0.9) = 3.4298. Use the results to show that (t) when t = 0.932.

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